/*
 * Copyright (C) 2008 The Android Open Source Project
 * 
 * Licensed under the Apache License, Version 2.0 (the "License"); you may not use this file except in compliance with the
 * License. You may obtain a copy of the License at
 * 
 * http://www.apache.org/licenses/LICENSE-2.0
 * 
 * Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an "AS IS"
 * BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the License for the specific language
 * governing permissions and limitations under the License.
 */

package org.oscim.utils;

import java.util.Arrays;
import java.util.Comparator;

/**
 * A stable, adaptive, iterative mergesort that requires far fewer than n lg(n)
 * comparisons when running on partially sorted
 * arrays, while offering performance comparable to a traditional mergesort when
 * run on random arrays. Like all proper mergesorts,
 * this sort is stable and runs O(n log n) time (worst case). In the worst case,
 * this sort requires temporary storage space for
 * n/2 object references; in the best case, it requires only a small constant
 * amount of space.
 * <p/>
 * This implementation was adapted from Tim Peters's list sort for Python, which
 * is described in detail here:
 * <p/>
 * http://svn.python.org/projects/python/trunk/Objects/listsort.txt
 * <p/>
 * Tim's C code may be found here:
 * <p/>
 * http://svn.python.org/projects/python/trunk/Objects/listobject.c
 * <p/>
 * The underlying techniques are described in this paper (and may have even
 * earlier origins):
 * <p/>
 * "Optimistic Sorting and Information Theoretic Complexity" Peter McIlroy SODA
 * (Fourth Annual ACM-SIAM Symposium on Discrete
 * Algorithms), pp 467-474, Austin, Texas, 25-27 January 1993.
 * <p/>
 * While the API to this class consists solely of static methods, it is
 * (privately) instantiable; a TimSort instance holds the
 * state of an ongoing sort, assuming the input array is large enough to warrant
 * the full-blown TimSort. Small arrays are sorted
 * in place, using a binary insertion sort.
 */
public class TimSort<T> {
    /**
     * This is the minimum sized sequence that will be merged. Shorter sequences
     * will be lengthened by calling binarySort. If the
     * entire array is less than this length, no merges will be performed.
     * <p/>
     * This constant should be a power of two. It was 64 in Tim Peter's C
     * implementation, but 32 was empirically determined to work
     * better in this implementation. In the unlikely event that you set this
     * constant to be a number that's not a power of two,
     * you'll need to change the {@link #minRunLength} computation.
     * <p/>
     * If you decrease this constant, you must change the stackLen computation
     * in the TimSort constructor, or you risk an
     * ArrayOutOfBounds exception. See listsort.txt for a discussion of the
     * minimum stack length required as a function of the
     * length of the array being sorted and the minimum merge sequence length.
     */
    private static final int MIN_MERGE = 32;

    /**
     * The array being sorted.
     */
    private T[] a;

    /**
     * The comparator for this sort.
     */
    private Comparator<? super T> c;

    /**
     * When we get into galloping mode, we stay there until both runs win less
     * often than MIN_GALLOP consecutive times.
     */
    private static final int MIN_GALLOP = 7;

    /**
     * This controls when we get *into* galloping mode. It is initialized to
     * MIN_GALLOP. The mergeLo and mergeHi methods nudge it
     * higher for random data, and lower for highly structured data.
     */
    private int minGallop = MIN_GALLOP;

    /**
     * Maximum initial size of tmp array, which is used for merging. The array
     * can grow to accommodate demand.
     * <p/>
     * Unlike Tim's original C version, we do not allocate this much storage
     * when sorting smaller arrays. This change was required
     * for performance.
     */
    private static final int INITIAL_TMP_STORAGE_LENGTH = 256;

    /**
     * Temp storage for merges.
     */
    private T[] tmp; // Actual runtime type will be Object[], regardless of T
    private int tmpCount;

    /**
     * A stack of pending runs yet to be merged. Run i starts at address base[i]
     * and extends for len[i] elements. It's always true
     * (so long as the indices are in bounds) that:
     * <p/>
     * runBase[i] + runLen[i] == runBase[i + 1]
     * <p/>
     * so we could cut the storage for this, but it's a minor amount, and
     * keeping all the info explicit simplifies the code.
     */
    private int stackSize = 0; // Number of pending runs on stack
    private final int[] runBase;
    private final int[] runLen;

    /**
     * Asserts have been placed in if-statements for performace. To enable them,
     * set this field to true and enable them in VM with
     * a command line flag. If you modify this class, please do test the
     * asserts!
     */
    private static final boolean DEBUG = false;

    @SuppressWarnings("unchecked")
    public TimSort() {
        tmp = (T[]) new Object[INITIAL_TMP_STORAGE_LENGTH];
        runBase = new int[40];
        runLen = new int[40];
    }

    public void doSort(T[] a, Comparator<T> c, int lo, int hi) {
        stackSize = 0;
        rangeCheck(a.length, lo, hi);
        int nRemaining = hi - lo;
        if (nRemaining < 2)
            return; // Arrays of size 0 and 1 are always sorted

        // If array is small, do a "mini-TimSort" with no merges
        if (nRemaining < MIN_MERGE) {
            int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
            binarySort(a, lo, hi, lo + initRunLen, c);
            return;
        }

        this.a = a;
        this.c = c;
        tmpCount = 0;

        /**
         * March over the array once, left to right, finding natural runs,
         * extending short natural runs to minRun elements, and
         * merging runs to maintain stack invariant.
         */
        int minRun = minRunLength(nRemaining);
        do {
            // Identify next run
            int runLen = countRunAndMakeAscending(a, lo, hi, c);

            // If run is short, extend to min(minRun, nRemaining)
            if (runLen < minRun) {
                int force = nRemaining <= minRun ? nRemaining : minRun;
                binarySort(a, lo, lo + force, lo + runLen, c);
                runLen = force;
            }

            // Push run onto pending-run stack, and maybe merge
            pushRun(lo, runLen);
            mergeCollapse();

            // Advance to find next run
            lo += runLen;
            nRemaining -= runLen;
        } while (nRemaining != 0);

        // Merge all remaining runs to complete sort
        if (DEBUG)
            assert lo == hi;
        mergeForceCollapse();
        if (DEBUG)
            assert stackSize == 1;

        this.a = null;
        this.c = null;
        T[] tmp = this.tmp;
        for (int i = 0, n = tmpCount; i < n; i++)
            tmp[i] = null;
    }

    /**
     * Creates a TimSort instance to maintain the state of an ongoing sort.
     *
     * @param a the array to be sorted
     * @param c the comparator to determine the order of the sort
     */
    private TimSort(T[] a, Comparator<? super T> c) {
        this.a = a;
        this.c = c;

        // Allocate temp storage (which may be increased later if necessary)
        int len = a.length;
        @SuppressWarnings("unchecked")
        T[] newArray = (T[]) new Object[len < 2 * INITIAL_TMP_STORAGE_LENGTH ? len >>> 1
                : INITIAL_TMP_STORAGE_LENGTH];
        tmp = newArray;

        /*
         * Allocate runs-to-be-merged stack (which cannot be expanded). The
         * stack length requirements are described in listsort.txt.
         * The C version always uses the same stack length (85), but this was
         * measured to be too expensive when sorting "mid-sized"
         * arrays (e.g., 100 elements) in Java. Therefore, we use smaller (but
         * sufficiently large) stack lengths for smaller arrays.
         * The "magic numbers" in the computation below must be changed if
         * MIN_MERGE is decreased. See the MIN_MERGE declaration
         * above for more information.
         */
        int stackLen = (len < 120 ? 5 : len < 1542 ? 10 : len < 119151 ? 19 : 40);
        runBase = new int[stackLen];
        runLen = new int[stackLen];
    }

    /*
     * The next two methods (which are package private and static) constitute
     * the entire API of this class. Each of these methods
     * obeys the contract of the public method with the same signature in
     * java.util.Arrays.
     */

    static <T> void sort(T[] a, Comparator<? super T> c) {
        sort(a, 0, a.length, c);
    }

    static <T> void sort(T[] a, int lo, int hi, Comparator<? super T> c) {
        if (c == null) {
            Arrays.sort(a, lo, hi);
            return;
        }

        rangeCheck(a.length, lo, hi);
        int nRemaining = hi - lo;
        if (nRemaining < 2)
            return; // Arrays of size 0 and 1 are always sorted

        // If array is small, do a "mini-TimSort" with no merges
        if (nRemaining < MIN_MERGE) {
            int initRunLen = countRunAndMakeAscending(a, lo, hi, c);
            binarySort(a, lo, hi, lo + initRunLen, c);
            return;
        }

        /**
         * March over the array once, left to right, finding natural runs,
         * extending short natural runs to minRun elements, and
         * merging runs to maintain stack invariant.
         */
        TimSort<T> ts = new TimSort<T>(a, c);
        int minRun = minRunLength(nRemaining);
        do {
            // Identify next run
            int runLen = countRunAndMakeAscending(a, lo, hi, c);

            // If run is short, extend to min(minRun, nRemaining)
            if (runLen < minRun) {
                int force = nRemaining <= minRun ? nRemaining : minRun;
                binarySort(a, lo, lo + force, lo + runLen, c);
                runLen = force;
            }

            // Push run onto pending-run stack, and maybe merge
            ts.pushRun(lo, runLen);
            ts.mergeCollapse();

            // Advance to find next run
            lo += runLen;
            nRemaining -= runLen;
        } while (nRemaining != 0);

        // Merge all remaining runs to complete sort
        if (DEBUG)
            assert lo == hi;
        ts.mergeForceCollapse();
        if (DEBUG)
            assert ts.stackSize == 1;
    }

    /**
     * Sorts the specified portion of the specified array using a binary
     * insertion sort. This is the best method for sorting small
     * numbers of elements. It requires O(n log n) compares, but O(n^2) data
     * movement (worst case).
     * <p/>
     * If the initial part of the specified range is already sorted, this method
     * can take advantage of it: the method assumes that
     * the elements from index {@code lo}, inclusive, to {@code start},
     * exclusive
     * are already sorted.
     *
     * @param a     the array in which a range is to be sorted
     * @param lo    the index of the first element in the range to be sorted
     * @param hi    the index after the last element in the range to be sorted
     * @param start the index of the first element in the range that is not
     *              already known to be sorted (@code lo <= start <= hi}
     * @param c     comparator to used for the sort
     */
    @SuppressWarnings("fallthrough")
    private static <T> void binarySort(T[] a, int lo, int hi, int start, Comparator<? super T> c) {
        if (DEBUG)
            assert lo <= start && start <= hi;
        if (start == lo)
            start++;
        for (; start < hi; start++) {
            T pivot = a[start];

            // Set left (and right) to the index where a[start] (pivot) belongs
            int left = lo;
            int right = start;
            if (DEBUG)
                assert left <= right;
            /*
             * Invariants: pivot >= all in [lo, left). pivot < all in [right,
             * start).
             */
            while (left < right) {
                int mid = (left + right) >>> 1;
                if (c.compare(pivot, a[mid]) < 0)
                    right = mid;
                else
                    left = mid + 1;
            }
            if (DEBUG)
                assert left == right;

            /*
             * The invariants still hold: pivot >= all in [lo, left) and pivot <
             * all in [left, start), so pivot belongs at left. Note
             * that if there are elements equal to pivot, left points to the
             * first slot after them -- that's why this sort is stable.
             * Slide elements over to make room to make room for pivot.
             */
            int n = start - left; // The number of elements to move
            // Switch is just an optimization for arraycopy in default case
            switch (n) {
                case 2:
                    a[left + 2] = a[left + 1];
                case 1:
                    a[left + 1] = a[left];
                    break;
                default:
                    System.arraycopy(a, left, a, left + 1, n);
            }
            a[left] = pivot;
        }
    }

    /**
     * Returns the length of the run beginning at the specified position in the
     * specified array and reverses the run if it is
     * descending (ensuring that the run will always be ascending when the
     * method returns).
     * <p/>
     * A run is the longest ascending sequence with:
     * <p/>
     * a[lo] <= a[lo + 1] <= a[lo + 2] <= ...
     * <p/>
     * or the longest descending sequence with:
     * <p/>
     * a[lo] > a[lo + 1] > a[lo + 2] > ...
     * <p/>
     * For its intended use in a stable mergesort, the strictness of the
     * definition of "descending" is needed so that the call can
     * safely reverse a descending sequence without violating stability.
     *
     * @param a  the array in which a run is to be counted and possibly reversed
     * @param lo index of the first element in the run
     * @param hi index after the last element that may be contained in the run.
     *           It is required that @code{lo < hi}.
     * @param c  the comparator to used for the sort
     * @return the length of the run beginning at the specified position in the
     * specified array
     */
    private static <T> int countRunAndMakeAscending(T[] a, int lo, int hi, Comparator<? super T> c) {
        if (DEBUG)
            assert lo < hi;
        int runHi = lo + 1;
        if (runHi == hi)
            return 1;

        // Find end of run, and reverse range if descending
        if (c.compare(a[runHi++], a[lo]) < 0) { // Descending
            while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) < 0)
                runHi++;
            reverseRange(a, lo, runHi);
        } else { // Ascending
            while (runHi < hi && c.compare(a[runHi], a[runHi - 1]) >= 0)
                runHi++;
        }

        return runHi - lo;
    }

    /**
     * Reverse the specified range of the specified array.
     *
     * @param a  the array in which a range is to be reversed
     * @param lo the index of the first element in the range to be reversed
     * @param hi the index after the last element in the range to be reversed
     */
    private static void reverseRange(Object[] a, int lo, int hi) {
        hi--;
        while (lo < hi) {
            Object t = a[lo];
            a[lo++] = a[hi];
            a[hi--] = t;
        }
    }

    /**
     * Returns the minimum acceptable run length for an array of the specified
     * length. Natural runs shorter than this will be
     * extended with {@link #binarySort}.
     * <p/>
     * Roughly speaking, the computation is:
     * <p/>
     * If n < MIN_MERGE, return n (it's too small to bother with fancy stuff).
     * Else if n is an exact power of 2, return
     * MIN_MERGE/2. Else return an int k, MIN_MERGE/2 <= k <= MIN_MERGE, such
     * that n/k is close to, but strictly less than, an
     * exact power of 2.
     * <p/>
     * For the rationale, see listsort.txt.
     *
     * @param n the length of the array to be sorted
     * @return the length of the minimum run to be merged
     */
    private static int minRunLength(int n) {
        if (DEBUG)
            assert n >= 0;
        int r = 0; // Becomes 1 if any 1 bits are shifted off
        while (n >= MIN_MERGE) {
            r |= (n & 1);
            n >>= 1;
        }
        return n + r;
    }

    /**
     * Pushes the specified run onto the pending-run stack.
     *
     * @param runBase index of the first element in the run
     * @param runLen  the number of elements in the run
     */
    private void pushRun(int runBase, int runLen) {
        this.runBase[stackSize] = runBase;
        this.runLen[stackSize] = runLen;
        stackSize++;
    }

    /**
     * Examines the stack of runs waiting to be merged and merges adjacent runs
     * until the stack invariants are reestablished:
     * <p/>
     * 1. runLen[i - 3] > runLen[i - 2] + runLen[i - 1] 2. runLen[i - 2] >
     * runLen[i - 1]
     * <p/>
     * This method is called each time a new run is pushed onto the stack, so
     * the invariants are guaranteed to hold for i <
     * stackSize upon entry to the method.
     */
    private void mergeCollapse() {
        while (stackSize > 1) {
            int n = stackSize - 2;
            if (n > 0 && runLen[n - 1] <= runLen[n] + runLen[n + 1]) {
                if (runLen[n - 1] < runLen[n + 1])
                    n--;
                mergeAt(n);
            } else if (runLen[n] <= runLen[n + 1]) {
                mergeAt(n);
            } else {
                break; // Invariant is established
            }
        }
    }

    /**
     * Merges all runs on the stack until only one remains. This method is
     * called once, to complete the sort.
     */
    private void mergeForceCollapse() {
        while (stackSize > 1) {
            int n = stackSize - 2;
            if (n > 0 && runLen[n - 1] < runLen[n + 1])
                n--;
            mergeAt(n);
        }
    }

    /**
     * Merges the two runs at stack indices i and i+1. Run i must be the
     * penultimate or antepenultimate run on the stack. In other
     * words, i must be equal to stackSize-2 or stackSize-3.
     *
     * @param i stack index of the first of the two runs to merge
     */
    private void mergeAt(int i) {
        if (DEBUG)
            assert stackSize >= 2;
        if (DEBUG)
            assert i >= 0;
        if (DEBUG)
            assert i == stackSize - 2 || i == stackSize - 3;

        int base1 = runBase[i];
        int len1 = runLen[i];
        int base2 = runBase[i + 1];
        int len2 = runLen[i + 1];
        if (DEBUG)
            assert len1 > 0 && len2 > 0;
        if (DEBUG)
            assert base1 + len1 == base2;

        /*
         * Record the length of the combined runs; if i is the 3rd-last run now,
         * also slide over the last run (which isn't involved
         * in this merge). The current run (i+1) goes away in any case.
         */
        runLen[i] = len1 + len2;
        if (i == stackSize - 3) {
            runBase[i + 1] = runBase[i + 2];
            runLen[i + 1] = runLen[i + 2];
        }
        stackSize--;

        /*
         * Find where the first element of run2 goes in run1. Prior elements in
         * run1 can be ignored (because they're already in
         * place).
         */
        int k = gallopRight(a[base2], a, base1, len1, 0, c);
        if (DEBUG)
            assert k >= 0;
        base1 += k;
        len1 -= k;
        if (len1 == 0)
            return;

        /*
         * Find where the last element of run1 goes in run2. Subsequent elements
         * in run2 can be ignored (because they're already in
         * place).
         */
        len2 = gallopLeft(a[base1 + len1 - 1], a, base2, len2, len2 - 1, c);
        if (DEBUG)
            assert len2 >= 0;
        if (len2 == 0)
            return;

        // Merge remaining runs, using tmp array with min(len1, len2) elements
        if (len1 <= len2)
            mergeLo(base1, len1, base2, len2);
        else
            mergeHi(base1, len1, base2, len2);
    }

    /**
     * Locates the position at which to insert the specified key into the
     * specified sorted range; if the range contains an element
     * equal to key, returns the index of the leftmost equal element.
     *
     * @param key  the key whose insertion point to search for
     * @param a    the array in which to search
     * @param base the index of the first element in the range
     * @param len  the length of the range; must be > 0
     * @param hint the index at which to begin the search, 0 <= hint < n. The
     *             closer hint is to the result, the faster this method
     *             will run.
     * @param c    the comparator used to order the range, and to search
     * @return the int k, 0 <= k <= n such that a[b + k - 1] < key <= a[b + k],
     * pretending that a[b - 1] is minus infinity and a[b
     * + n] is infinity. In other words, key belongs at index b + k; or
     * in other words, the first k elements of a should
     * precede key, and the last n - k should follow it.
     */
    private static <T> int gallopLeft(T key, T[] a, int base, int len, int hint,
                                      Comparator<? super T> c) {
        if (DEBUG)
            assert len > 0 && hint >= 0 && hint < len;
        int lastOfs = 0;
        int ofs = 1;
        if (c.compare(key, a[base + hint]) > 0) {
            // Gallop right until a[base+hint+lastOfs] < key <= a[base+hint+ofs]
            int maxOfs = len - hint;
            while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) > 0) {
                lastOfs = ofs;
                ofs = (ofs << 1) + 1;
                if (ofs <= 0) // int overflow
                    ofs = maxOfs;
            }
            if (ofs > maxOfs)
                ofs = maxOfs;

            // Make offsets relative to base
            lastOfs += hint;
            ofs += hint;
        } else { // key <= a[base + hint]
            // Gallop left until a[base+hint-ofs] < key <= a[base+hint-lastOfs]
            final int maxOfs = hint + 1;
            while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) <= 0) {
                lastOfs = ofs;
                ofs = (ofs << 1) + 1;
                if (ofs <= 0) // int overflow
                    ofs = maxOfs;
            }
            if (ofs > maxOfs)
                ofs = maxOfs;

            // Make offsets relative to base
            int tmp = lastOfs;
            lastOfs = hint - ofs;
            ofs = hint - tmp;
        }
        if (DEBUG)
            assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

        /*
         * Now a[base+lastOfs] < key <= a[base+ofs], so key belongs somewhere to
         * the right of lastOfs but no farther right than ofs.
         * Do a binary search, with invariant a[base + lastOfs - 1] < key <=
         * a[base + ofs].
         */
        lastOfs++;
        while (lastOfs < ofs) {
            int m = lastOfs + ((ofs - lastOfs) >>> 1);

            if (c.compare(key, a[base + m]) > 0)
                lastOfs = m + 1; // a[base + m] < key
            else
                ofs = m; // key <= a[base + m]
        }
        if (DEBUG)
            assert lastOfs == ofs; // so a[base + ofs - 1] < key <= a[base + ofs]
        return ofs;
    }

    /**
     * Like gallopLeft, except that if the range contains an element equal to
     * key, gallopRight returns the index after the
     * rightmost equal element.
     *
     * @param key  the key whose insertion point to search for
     * @param a    the array in which to search
     * @param base the index of the first element in the range
     * @param len  the length of the range; must be > 0
     * @param hint the index at which to begin the search, 0 <= hint < n. The
     *             closer hint is to the result, the faster this method
     *             will run.
     * @param c    the comparator used to order the range, and to search
     * @return the int k, 0 <= k <= n such that a[b + k - 1] <= key < a[b + k]
     */
    private static <T> int gallopRight(T key, T[] a, int base, int len, int hint,
                                       Comparator<? super T> c) {
        if (DEBUG)
            assert len > 0 && hint >= 0 && hint < len;

        int ofs = 1;
        int lastOfs = 0;
        if (c.compare(key, a[base + hint]) < 0) {
            // Gallop left until a[b+hint - ofs] <= key < a[b+hint - lastOfs]
            int maxOfs = hint + 1;
            while (ofs < maxOfs && c.compare(key, a[base + hint - ofs]) < 0) {
                lastOfs = ofs;
                ofs = (ofs << 1) + 1;
                if (ofs <= 0) // int overflow
                    ofs = maxOfs;
            }
            if (ofs > maxOfs)
                ofs = maxOfs;

            // Make offsets relative to b
            int tmp = lastOfs;
            lastOfs = hint - ofs;
            ofs = hint - tmp;
        } else { // a[b + hint] <= key
            // Gallop right until a[b+hint + lastOfs] <= key < a[b+hint + ofs]
            int maxOfs = len - hint;
            while (ofs < maxOfs && c.compare(key, a[base + hint + ofs]) >= 0) {
                lastOfs = ofs;
                ofs = (ofs << 1) + 1;
                if (ofs <= 0) // int overflow
                    ofs = maxOfs;
            }
            if (ofs > maxOfs)
                ofs = maxOfs;

            // Make offsets relative to b
            lastOfs += hint;
            ofs += hint;
        }
        if (DEBUG)
            assert -1 <= lastOfs && lastOfs < ofs && ofs <= len;

        /*
         * Now a[b + lastOfs] <= key < a[b + ofs], so key belongs somewhere to
         * the right of lastOfs but no farther right than ofs.
         * Do a binary search, with invariant a[b + lastOfs - 1] <= key < a[b +
         * ofs].
         */
        lastOfs++;
        while (lastOfs < ofs) {
            int m = lastOfs + ((ofs - lastOfs) >>> 1);

            if (c.compare(key, a[base + m]) < 0)
                ofs = m; // key < a[b + m]
            else
                lastOfs = m + 1; // a[b + m] <= key
        }
        if (DEBUG)
            assert lastOfs == ofs; // so a[b + ofs - 1] <= key < a[b + ofs]
        return ofs;
    }

    /**
     * Merges two adjacent runs in place, in a stable fashion. The first element
     * of the first run must be greater than the first
     * element of the second run (a[base1] > a[base2]), and the last element of
     * the first run (a[base1 + len1-1]) must be greater
     * than all elements of the second run.
     * <p/>
     * For performance, this method should be called only when len1 <= len2; its
     * twin, mergeHi should be called if len1 >= len2.
     * (Either method may be called if len1 == len2.)
     *
     * @param base1 index of first element in first run to be merged
     * @param len1  length of first run to be merged (must be > 0)
     * @param base2 index of first element in second run to be merged (must be
     *              aBase + aLen)
     * @param len2  length of second run to be merged (must be > 0)
     */
    private void mergeLo(int base1, int len1, int base2, int len2) {
        if (DEBUG)
            assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

        // Copy first run into temp array
        T[] a = this.a; // For performance
        T[] tmp = ensureCapacity(len1);
        System.arraycopy(a, base1, tmp, 0, len1);

        int cursor1 = 0; // Indexes into tmp array
        int cursor2 = base2; // Indexes int a
        int dest = base1; // Indexes int a

        // Move first element of second run and deal with degenerate cases
        a[dest++] = a[cursor2++];
        if (--len2 == 0) {
            System.arraycopy(tmp, cursor1, a, dest, len1);
            return;
        }
        if (len1 == 1) {
            System.arraycopy(a, cursor2, a, dest, len2);
            a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
            return;
        }

        Comparator<? super T> c = this.c; // Use local variable for performance
        int minGallop = this.minGallop; // "    " "     " "
        outer:
        while (true) {
            int count1 = 0; // Number of times in a row that first run won
            int count2 = 0; // Number of times in a row that second run won

            /*
             * Do the straightforward thing until (if ever) one run starts
             * winning consistently.
             */
            do {
                if (DEBUG)
                    assert len1 > 1 && len2 > 0;
                if (c.compare(a[cursor2], tmp[cursor1]) < 0) {
                    a[dest++] = a[cursor2++];
                    count2++;
                    count1 = 0;
                    if (--len2 == 0)
                        break outer;
                } else {
                    a[dest++] = tmp[cursor1++];
                    count1++;
                    count2 = 0;
                    if (--len1 == 1)
                        break outer;
                }
            } while ((count1 | count2) < minGallop);

            /*
             * One run is winning so consistently that galloping may be a huge
             * win. So try that, and continue galloping until (if
             * ever) neither run appears to be winning consistently anymore.
             */
            do {
                if (DEBUG)
                    assert len1 > 1 && len2 > 0;
                count1 = gallopRight(a[cursor2], tmp, cursor1, len1, 0, c);
                if (count1 != 0) {
                    System.arraycopy(tmp, cursor1, a, dest, count1);
                    dest += count1;
                    cursor1 += count1;
                    len1 -= count1;
                    if (len1 <= 1) // len1 == 1 || len1 == 0
                        break outer;
                }
                a[dest++] = a[cursor2++];
                if (--len2 == 0)
                    break outer;

                count2 = gallopLeft(tmp[cursor1], a, cursor2, len2, 0, c);
                if (count2 != 0) {
                    System.arraycopy(a, cursor2, a, dest, count2);
                    dest += count2;
                    cursor2 += count2;
                    len2 -= count2;
                    if (len2 == 0)
                        break outer;
                }
                a[dest++] = tmp[cursor1++];
                if (--len1 == 1)
                    break outer;
                minGallop--;
            } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
            if (minGallop < 0)
                minGallop = 0;
            minGallop += 2; // Penalize for leaving gallop mode
        } // End of "outer" loop
        this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field

        if (len1 == 1) {
            if (DEBUG)
                assert len2 > 0;
            System.arraycopy(a, cursor2, a, dest, len2);
            a[dest + len2] = tmp[cursor1]; // Last elt of run 1 to end of merge
        } else if (len1 == 0) {
            throw new IllegalArgumentException("Comparison method violates its general contract!");
        } else {
            if (DEBUG)
                assert len2 == 0;
            if (DEBUG)
                assert len1 > 1;
            System.arraycopy(tmp, cursor1, a, dest, len1);
        }
    }

    /**
     * Like mergeLo, except that this method should be called only if len1 >=
     * len2; mergeLo should be called if len1 <= len2.
     * (Either method may be called if len1 == len2.)
     *
     * @param base1 index of first element in first run to be merged
     * @param len1  length of first run to be merged (must be > 0)
     * @param base2 index of first element in second run to be merged (must be
     *              aBase + aLen)
     * @param len2  length of second run to be merged (must be > 0)
     */
    private void mergeHi(int base1, int len1, int base2, int len2) {
        if (DEBUG)
            assert len1 > 0 && len2 > 0 && base1 + len1 == base2;

        // Copy second run into temp array
        T[] a = this.a; // For performance
        T[] tmp = ensureCapacity(len2);
        System.arraycopy(a, base2, tmp, 0, len2);

        int cursor1 = base1 + len1 - 1; // Indexes into a
        int cursor2 = len2 - 1; // Indexes into tmp array
        int dest = base2 + len2 - 1; // Indexes into a

        // Move last element of first run and deal with degenerate cases
        a[dest--] = a[cursor1--];
        if (--len1 == 0) {
            System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
            return;
        }
        if (len2 == 1) {
            dest -= len1;
            cursor1 -= len1;
            System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
            a[dest] = tmp[cursor2];
            return;
        }

        Comparator<? super T> c = this.c; // Use local variable for performance
        int minGallop = this.minGallop; // "    " "     " "
        outer:
        while (true) {
            int count1 = 0; // Number of times in a row that first run won
            int count2 = 0; // Number of times in a row that second run won

            /*
             * Do the straightforward thing until (if ever) one run appears to
             * win consistently.
             */
            do {
                if (DEBUG)
                    assert len1 > 0 && len2 > 1;
                if (c.compare(tmp[cursor2], a[cursor1]) < 0) {
                    a[dest--] = a[cursor1--];
                    count1++;
                    count2 = 0;
                    if (--len1 == 0)
                        break outer;
                } else {
                    a[dest--] = tmp[cursor2--];
                    count2++;
                    count1 = 0;
                    if (--len2 == 1)
                        break outer;
                }
            } while ((count1 | count2) < minGallop);

            /*
             * One run is winning so consistently that galloping may be a huge
             * win. So try that, and continue galloping until (if
             * ever) neither run appears to be winning consistently anymore.
             */
            do {
                if (DEBUG)
                    assert len1 > 0 && len2 > 1;
                count1 = len1 - gallopRight(tmp[cursor2], a, base1, len1, len1 - 1, c);
                if (count1 != 0) {
                    dest -= count1;
                    cursor1 -= count1;
                    len1 -= count1;
                    System.arraycopy(a, cursor1 + 1, a, dest + 1, count1);
                    if (len1 == 0)
                        break outer;
                }
                a[dest--] = tmp[cursor2--];
                if (--len2 == 1)
                    break outer;

                count2 = len2 - gallopLeft(a[cursor1], tmp, 0, len2, len2 - 1, c);
                if (count2 != 0) {
                    dest -= count2;
                    cursor2 -= count2;
                    len2 -= count2;
                    System.arraycopy(tmp, cursor2 + 1, a, dest + 1, count2);
                    if (len2 <= 1) // len2 == 1 || len2 == 0
                        break outer;
                }
                a[dest--] = a[cursor1--];
                if (--len1 == 0)
                    break outer;
                minGallop--;
            } while (count1 >= MIN_GALLOP | count2 >= MIN_GALLOP);
            if (minGallop < 0)
                minGallop = 0;
            minGallop += 2; // Penalize for leaving gallop mode
        } // End of "outer" loop
        this.minGallop = minGallop < 1 ? 1 : minGallop; // Write back to field

        if (len2 == 1) {
            if (DEBUG)
                assert len1 > 0;
            dest -= len1;
            cursor1 -= len1;
            System.arraycopy(a, cursor1 + 1, a, dest + 1, len1);
            a[dest] = tmp[cursor2]; // Move first elt of run2 to front of merge
        } else if (len2 == 0) {
            throw new IllegalArgumentException("Comparison method violates its general contract!");
        } else {
            if (DEBUG)
                assert len1 == 0;
            if (DEBUG)
                assert len2 > 0;
            System.arraycopy(tmp, 0, a, dest - (len2 - 1), len2);
        }
    }

    /**
     * Ensures that the external array tmp has at least the specified number of
     * elements, increasing its size if necessary. The
     * size increases exponentially to ensure amortized linear time complexity.
     *
     * @param minCapacity the minimum required capacity of the tmp array
     * @return tmp, whether or not it grew
     */
    private T[] ensureCapacity(int minCapacity) {
        tmpCount = Math.max(tmpCount, minCapacity);
        if (tmp.length < minCapacity) {
            // Compute smallest power of 2 > minCapacity
            int newSize = minCapacity;
            newSize |= newSize >> 1;
            newSize |= newSize >> 2;
            newSize |= newSize >> 4;
            newSize |= newSize >> 8;
            newSize |= newSize >> 16;
            newSize++;

            if (newSize < 0) // Not bloody likely!
                newSize = minCapacity;
            else
                newSize = Math.min(newSize, a.length >>> 1);

            @SuppressWarnings("unchecked")
            T[] newArray = (T[]) new Object[newSize];
            tmp = newArray;
        }
        return tmp;
    }

    /**
     * Checks that fromIndex and toIndex are in range, and throws an appropriate
     * exception if they aren't.
     *
     * @param arrayLen  the length of the array
     * @param fromIndex the index of the first element of the range
     * @param toIndex   the index after the last element of the range
     * @throws IllegalArgumentException       if fromIndex > toIndex
     * @throws ArrayIndexOutOfBoundsException if fromIndex < 0 or toIndex >
     *                                        arrayLen
     */
    private static void rangeCheck(int arrayLen, int fromIndex, int toIndex) {
        if (fromIndex > toIndex)
            throw new IllegalArgumentException("fromIndex(" + fromIndex + ") > toIndex(" + toIndex
                    + ")");
        if (fromIndex < 0)
            throw new ArrayIndexOutOfBoundsException(fromIndex);
        if (toIndex > arrayLen)
            throw new ArrayIndexOutOfBoundsException(toIndex);
    }
}
